Use the intermediate value theorem to approximate the one real zero on the given interval. Round your answer to two
decimal places.
-$$f ( x ) = x ^ { 3 } - 8 x - 3 ; [ - 3 , - 2 ]$$
A) $- 2.62$
B) $- 0.38$
C) $0.38$
D) $2.62$

Question

Quizplus · Accepted Answer

The Intermediate Value Theorem states that if $f$ is continuous on a closed interval $[a, b]$ and $N$ is any number between $f(a)$ and $f(b)$, then there exists at least one number $c$ in the interval $[a, b]$ such that $f(c) = N$. To find the zero of $f(x) = x^3 - 8x - 3$ in the interval $[-3, -2]$, we evaluate $f(-3)$ and $f(-2)$:- $f(-3) = (-3)^3 - 8(-3) - 3 = -27 + 24 - 3 = -6$- $f(-2) = (-2)^3 - 8(-2) - 3 = -8 + 16 - 3 = 5$Since $f(-3)$ is negative and $f(-2)$ is positive, and $f(x)$ is continuous (being a polynomial), there must be at least one real zero in the interval $[-3, -2]$ by the Intermediate Value Theorem. The correct answer, rounded to two decimal places, is $-2.62$, which is a reasonable approximation of the zero in this interval.

Use the Intermediate Value Theorem to Approximate the One Real f(x)=x3−8x−3;[−3,−2]f ( x ) = x ^ { 3 } - 8 x - 3 ; [ - 3 , - 2 ]f(x)=x3−8x−3;[−3,−2]

Use the Intermediate Value Theorem to Approximate the One Real $f ( x ) = x ^ { 3 } - 8 x - 3 ; [ - 3 , - 2 ]$